Palindrome Pairs -- LeetCode 336

Given a list of unique words. Find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome.

Example 1:
Given words = ["bat", "tab", "cat"]
Return [[0, 1], [1, 0]]
The palindromes are ["battab", "tabbat"]

Example 2:
Given words = ["abcd", "dcba", "lls", "s", "sssll"]
Return [[0, 1], [1, 0], [3, 2], [2, 4]]
The palindromes are ["dcbaabcd", "abcddcba", "slls", "llssssll"]

Links: https://leetcode.com/discuss/93599/easy-to-understand-ac-c-solution-o-n-k-2-using-map

Solution 1:

class Solution {
public:
    vector<vector<int>> palindromePairs(vector<string>& words) {
        unordered_map<string, int> dict;
        vector<vector<int>> ans;
        // build dictionary
        for(int i = ; i < words.size(); i++) {
            string key = words[i];
            reverse(key.begin(), key.end());
            dict[key] = i;
        }
        // edge case: if empty string "" exists, find all palindromes to become pairs ("", self)
        if(dict.find("")!=dict.end()){
            for(int i = ; i < words.size(); i++){
                if(i == dict[""]) continue;
                if(isPalindrome(words[i])) ans.push_back({dict[""], i});
            }
        }

        for(int i = ; i < words.size(); i++) {
            for(int j = ; j < words[i].size(); j++) {
                string left = words[i].substr(, j);
                string right = words[i].substr(j, words[i].size() - j);

                if(dict.find(left) != dict.end() && isPalindrome(right) 
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　&& dict[left] != i) {
                    ans.push_back({i, dict[left]});
                }

                if(dict.find(right) != dict.end() && isPalindrome(left) 
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　&& dict[right] != i) {
                    ans.push_back({dict[right], i});
                }
            }
        }

        return ans;
    }

    bool isPalindrome(string str){
        int i = ;
        int j = str.size() - ; 

        while(i < j) {
            if(str[i++] != str[j--]) return false;
        }

        return true;
    }

};

Solution 2: